Based on Heisenberg's uncertainty principle,the uncertainty in the velocity of the electron to be found within an atomic nucleus of diameter $10^{-15} \ m$ is ............. $\times 10^9 \ ms^{-1}$ (nearest integer)
[Given : mass of electron $= 9.1 \times 10^{-31} \ kg$,Plank's constant $(h) = 6.626 \times 10^{-34} \ Js$ ]
(Value of $\pi = 3.14$ )

The equation $\Delta x \cdot \Delta p \ge \frac{h}{4\pi}$ shows

In an atom,an electron is moving with a speed of $600 \, m/s$ with an accuracy of $0.005 \%$. The certainty with which the position of the electron can be located is $(h = 6.6 \times 10^{-34} \, kg \, m^2 s^{-1}, m_e = 9.1 \times 10^{-31} \, kg)$:

If the uncertainty in velocity of electron $(\Delta v)$ is $0.1 \ m/s$,the uncertainty in its position $(\Delta x)$ is (given: $m_e = 9.1 \times 10^{-31} \ kg$)

Given below are two statements $:$
Statement $(I):$ It is impossible to specify simultaneously with arbitrary precision,both the linear momentum and the position of a particle.
Statement $(II) :$ If the uncertainty in the measurement of position and uncertainty in measurement of momentum are equal for an electron,then the uncertainty in the measurement of velocity is $\geq \sqrt{\frac{h}{4\pi}} \times \frac{1}{m}$ which simplifies to $\geq \frac{1}{2m} \sqrt{\frac{h}{\pi}}$. In the light of the above statements,choose the correct answer from the options given below $:$

"The position and velocity of a small particle like an electron cannot be simultaneously determined." This statement is: